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The area formed by triangular shaped region bounded by the curves $y=\sin x, y=\cos x$ and $x=0$ is
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The correct answer is:
$\sqrt{2}-1$
Given required area has been shown in the figure.
$x=\frac{\pi}{4}$ is the point of intersection of both curve
$\begin{aligned} \therefore \text { Required area } & =\int_0^{\pi / 4}(\cos x-\sin x) d x \\ & =[\sin x+\cos x]_0^{\pi / 4}=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right] \\ & =\frac{2}{\sqrt{2}}-1=\sqrt{2}-1 .\end{aligned}$
$x=\frac{\pi}{4}$ is the point of intersection of both curve
$\begin{aligned} \therefore \text { Required area } & =\int_0^{\pi / 4}(\cos x-\sin x) d x \\ & =[\sin x+\cos x]_0^{\pi / 4}=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right] \\ & =\frac{2}{\sqrt{2}}-1=\sqrt{2}-1 .\end{aligned}$
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